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alexdok [17]
3 years ago
9

Tablets are on sale for 15% off original price with the function p(t)=0.85t

Mathematics
1 answer:
Len [333]3 years ago
4 0
The answer would be 45% there you go
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Find the total surface area of the following cylinder: r=3 cm 3 cm SA = [? ]n cm2
OLga [1]

Answer:

<u>36</u>

Step-by-step explanation:

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2 years ago
Help me please <br> Select the correct answer.<br> What is the product of
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Answer: It's A. -7.

Step-by-step explanation: I've attached an Image of my work.

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2 years ago
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Question 2: You want to donate 30% of the $400 earned at your yard sale to
lilavasa [31]

Answer:

10%= 40$ 30%=120$

Step-by-step explanation:

ok so u earn 400$ at a yard sale. you divide it by 100 400/100=4. ok so then you multiply four by 10=40 and that is 10% multiply it by 30 u get 120 or 30%

sorry for the bad spelling

4 0
3 years ago
6 plates and 5 cups cost 32.10 dollars. 7 plates and 6 cups cost 37.70 dollars. Each plate costs the same amount as the other pl
slavikrds [6]

Answer:

Plates are 2.80 Dollars and cups are 1.40 dollars

Step-by-step explanation:

Cups = c

Plates = p

6p + 5c = 23.80 ==> multiply by 6 ==> 36p + 30c = 142.80

7p + 6c = 28.00 ==> multiply by 5 ==> 35p + 30c = 140.00

Now use algebra to have the 30c be on one side and the rest on the other:

36p + 30c = 142.80 | -36p

30c = 142.80 - 36p

35p + 30c = 140.00 | -35p

30c = 140.00 - 35p

Now set them equal to each other about the 30c:

142.80 - 36p = 140.00 - 35p

Use algebra to solve for p:

142.80 - 36p = 140.00 - 35p | +36p

142.80 = 140.00 + p | - 140

2.80 = p

Go back to one of the "30c" equations and plug in the value for p:

30c = 140 - 35p

30c = 140 - 35(2.80)

30c = 140 - 98

30c = 42 | /30

c = 42/30

c = 1.4

8 0
3 years ago
Read 2 more answers
Enter the coefficients of the fifth Taylor polynomial T5(x) for the function f(x) = x5−3x4+2x2+5x−2 based at b=1. T5(x)= + (x−1)
DENIUS [597]

Compute the necessary values/derivatives of f(x) at x=1:

f(1)=3

f'(1)=2

f''(1)=-12

f'''(1)=-12

f^{(4)}(1)=48

f^{(5)}(1)=120

Taylor's theorem then says we can "approximate" (in quotes because the Taylor polynomial for a polynomial is another, exact polynomial) f(x) at x=1 by

T_5(x)=\dfrac3{0!}+\dfrac2{1!}(x-1)-\dfrac{12}{2!}(x-1)^2-\dfrac{12}{3!}(x-1)^3+\dfrac{48}{4!}(x-1)^4+\dfrac{120}{5!}(x-1)^5

T_5(x)=3+2(x-1)-6(x-1)^2-2(x-1)^3+2(x-1)^4+(x-1)^5

###

Another way of doing this would be to solve for the coefficients a,b,c,d,e,g in

f(x)=a+b(x-1)+c(x-1)^2+d(x-1)^3+e(x-1)^4+g(x-1)^5

by expanding the right hand side and matching up terms with the same power of x.

5 0
2 years ago
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